+0  
 
0
315
2
avatar

cos(atan(sqrt(3))+asin(1/3)

Guest May 7, 2015

Best Answer 

 #1
avatar+18715 
+10

cos( atan( sqrt(3) )+asin(1/3) ) ?

$$\boxed{
\mathbf{
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)}
~\right)
}
} =~ ?
}$$

 

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\cos{ \left(\alpha_{rad} + \beta_{rad} \right) }
= \cos{ \left(\alpha_{rad} \right) } \cdot
\cos{ \left(\beta_{rad} \right) }
- \sin{ \left(\alpha_{rad} \right) } \cdot
\sin{ \left(\beta_{rad} \right) }
} & \quad \mathbf{
\alpha_{rad} = \arctan{(a)} \quad a= \sqrt{3} } \\
& \mathbf{
\quad \beta_{rad} = \arcsin{(b)} \quad b= \dfrac{1}{3} }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot
\sin{ \left(\arcsin{(b)} \right) } \\
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot b
\end{array}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{1}{\sqrt{1+a^2}}
=\pm\dfrac{1}{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{1}{2}
}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arcsin{(b)} ~\right) }
=\pm\sqrt{1-b^2}
=\pm \sqrt{1+ \left( \dfrac{1}{3} \right)^2}
=\pm \dfrac{\sqrt{8} }{ 3 }
}
$}}$$

$$\small{\text{
$
\boxed{
\sin{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{a}{\sqrt{1+a^2}}
=\pm\dfrac{ \sqrt{3} }{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{ \sqrt{3} }{2}
}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\left( \pm \dfrac{1}{2} \right) \cdot
\left( \pm \dfrac{ \sqrt{8} }{ 3 } \right)
-\left( \pm \dfrac{ \sqrt{3} }{ 2 } \right) \cdot
\left( \dfrac{1}{3} \right) \\
&=& \dfrac{1}{6} \left( \pm \sqrt{8} \pm\sqrt{3} \right)
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)} ~\right)
}
&=& \dfrac{1}{6} \left( + \sqrt{8} + \sqrt{3} \right)
= 0.76007965539 \\\\
&=& \dfrac{1}{6} \left( + \sqrt{8} - \sqrt{3} \right)
= 0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} + \sqrt{3} \right)
= -0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} - \sqrt{3} \right)
= -0.76007965539
\end{array}
$}}$$

 

 

heureka  May 8, 2015
Sort: 

2+0 Answers

 #1
avatar+18715 
+10
Best Answer

cos( atan( sqrt(3) )+asin(1/3) ) ?

$$\boxed{
\mathbf{
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)}
~\right)
}
} =~ ?
}$$

 

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\cos{ \left(\alpha_{rad} + \beta_{rad} \right) }
= \cos{ \left(\alpha_{rad} \right) } \cdot
\cos{ \left(\beta_{rad} \right) }
- \sin{ \left(\alpha_{rad} \right) } \cdot
\sin{ \left(\beta_{rad} \right) }
} & \quad \mathbf{
\alpha_{rad} = \arctan{(a)} \quad a= \sqrt{3} } \\
& \mathbf{
\quad \beta_{rad} = \arcsin{(b)} \quad b= \dfrac{1}{3} }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot
\sin{ \left(\arcsin{(b)} \right) } \\
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot b
\end{array}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{1}{\sqrt{1+a^2}}
=\pm\dfrac{1}{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{1}{2}
}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arcsin{(b)} ~\right) }
=\pm\sqrt{1-b^2}
=\pm \sqrt{1+ \left( \dfrac{1}{3} \right)^2}
=\pm \dfrac{\sqrt{8} }{ 3 }
}
$}}$$

$$\small{\text{
$
\boxed{
\sin{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{a}{\sqrt{1+a^2}}
=\pm\dfrac{ \sqrt{3} }{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{ \sqrt{3} }{2}
}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\left( \pm \dfrac{1}{2} \right) \cdot
\left( \pm \dfrac{ \sqrt{8} }{ 3 } \right)
-\left( \pm \dfrac{ \sqrt{3} }{ 2 } \right) \cdot
\left( \dfrac{1}{3} \right) \\
&=& \dfrac{1}{6} \left( \pm \sqrt{8} \pm\sqrt{3} \right)
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)} ~\right)
}
&=& \dfrac{1}{6} \left( + \sqrt{8} + \sqrt{3} \right)
= 0.76007965539 \\\\
&=& \dfrac{1}{6} \left( + \sqrt{8} - \sqrt{3} \right)
= 0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} + \sqrt{3} \right)
= -0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} - \sqrt{3} \right)
= -0.76007965539
\end{array}
$}}$$

 

 

heureka  May 8, 2015
 #2
avatar+91053 
0

That looks very impressive Heureka :))

Melody  May 8, 2015

6 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details